The Triangle Solver
Law of Sines, Law of Cosines, and area of oblique triangles
Step 1 of 9
Every triangle you have solved so far had a right angle. Most real triangles do not.
This triangle has no right angle. Your right-triangle tools do not apply here. So we build new ones -- out of the old ones.
Look at this triangle. Can you use SOH CAH TOA directly on it?
Step 2 of 9
Drop an altitude from vertex B to side b. Two right triangles appear.
Now you have right angles. The altitude \(h\) splits the oblique triangle into two right triangles. Use what you know.
Step 3 of 9
Write the altitude two ways. In the left right triangle, the altitude is opposite angle A. In the right triangle, it is opposite angle C.
From the left triangle, using \(\sin A\):
Step 4 of 9
Same segment. Two expressions. Set them equal: \(c \sin A = a \sin C\).
Rearrange this equation. Divide both sides by \(\sin A \cdot \sin C\). What do you get?
Step 5 of 9
The Law of Sines needs a matched angle-side pair. Which of these given-information patterns provides that?
Select all cases where the Law of Sines can start working:
Step 6 of 9
SSA: two sides and a non-included angle. Given \(a = 8\), \(b = 10\), \(A = 40°\). How many triangles fit these measurements?
Step 7 of 9
Explore the SSA boundary by dragging the side length \(a\). Watch what happens as you change it.
Drag the slider to discover the four SSA cases: too short (no triangle), exactly tangent (one right triangle), two crossings (two triangles), or longer than the other side (one triangle).
Step 8 of 9
SAS and SSS. Try the Law of Sines on SAS: you know two sides and the included angle. Can you find a matched pair?
Step 9 of 9
Area of a triangle: you know the base and you need the height. You already found \(h = c \sin A\). So what is the area?
Explore: The Triangle Solver
Drag any vertex to reshape the triangle. All measurements update live.
Mark known parts:
Drag a vertex or mark known parts to see which law applies and why.
Try This
The Fire
Two fire lookout towers sit 15 miles apart on an east-west line. Tower A (west) sights the smoke at bearing 038°. Tower B (east) sights it at bearing 312°. How far is the fire from each tower? Which crew is closer?
Enter the distance from Tower A to the fire (within 0.5 miles):
The Land Plot
A plot of land has four corners. Surveyor measurements give two triangles:
- Triangle 1: sides 120 ft, 90 ft, included angle 72°
- Triangle 2: sides 90 ft, 105 ft, included angle 85°
Find the total area. (Within 50 sq ft.)
Cell Tower Triangulation
Two cell towers are 8 miles apart. Signal timing puts your phone 5 miles from Tower A and 6 miles from Tower B. How many possible positions does that give you?
- Derive the Law of Sines from the altitude of an oblique triangle
- Solve triangles using ASA, AAS, and SSA given information
- Identify and resolve the ambiguous case (SSA) including all four boundary cases
- Derive the Law of Cosines and recognize it as a generalization of the Pythagorean theorem
- Solve triangles using SAS and SSS given information
- Calculate the area of an oblique triangle using \(\frac{1}{2}ab\sin C\)
- Choose the correct law based on given information
Quick Check
You know sides \(a = 7\), \(b = 10\), and the included angle \(C = 50°\). Which approach should you use?
Instructor Notes
Teaching Notes
This simulation unifies the Law of Sines, Law of Cosines, and the triangle area formula through a single geometric move: dropping an altitude. The derivation in steps 1-4 mirrors the textbook approach but makes it interactive. Step 6 is the centerpiece -- the SSA ambiguous case is one of the most commonly tested topics and one students struggle with most.
The altitude-based derivation in step 8 (Law of Cosines) is algebraically heavy. The scaffolding breaks it into manageable pieces, but students may need to see the full derivation written out. The key insight to emphasize: when \(C = 90°\), \(\cos C = 0\) and the formula reduces to the Pythagorean theorem.
Common Student Errors
- Reaching for the Law of Sines when given SAS -- no matched pair exists
- Finding one SSA solution and stopping, missing the second triangle
- Forgetting to convert bearings to triangle angles in applied problems
- Using degrees in trig functions instead of radians (or vice versa)
- Confusing the area formula \(\frac{1}{2}ab\sin C\) with the Law of Cosines
Discussion Questions
- Why does SSA sometimes give two triangles but ASA never does?
- In the fire lookout problem, could you solve it if both towers measured the same bearing? Why or why not?
- When does the Law of Cosines reduce to the Pythagorean theorem, and what does that tell you about the relationship between the two?
- GPS needs a minimum of four satellites. We showed that two towers give two positions and three towers resolve it. What does the fourth satellite add?
Exam Connection
The SSA ambiguous case appears frequently on exams. Students should be able to: (1) identify when SSA applies, (2) determine how many triangles exist, (3) solve both triangles when two exist. The "which law?" decision is also a common exam question -- given a set of known parts, choose the correct approach.